Question

Which of the following compounds is formed as major product in the following reaction?
\[ \text{2-Methylbut-2-ene} \xrightarrow{\text{HBr, Peroxide}} \text{Product} \]

• A. 3-Bromo-2-methylbutane
• B. 2-Bromo-2-methylbutane
• C. 2-Bromo-3-methylbutane
• D. 2-Bromobutane

Hint:

In anti-Markovnikov addition reactions, the halogen adds to the less substituted carbon, and the hydrogen adds to the more substituted carbon.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Addition of \( \text{HBr} \) to an unsymmetrical alkene in the presence of peroxide follows the Anti-Markovnikov rule (Kharasch effect).
Step 2: Key Formula or Approach:
In Anti-Markovnikov addition, the bromine atom (\( \text{Br} \)) attaches to the doubly bonded carbon atom that has more hydrogen atoms.
Step 3: Detailed Explanation:
Structure of 2-Methylbut-2-ene: \( \text{CH}_3 - \text{C}(\text{CH}_3) = \text{CH} - \text{CH}_3 \).
- C2 has zero hydrogen atoms (it's attached to two methyls and the double bond).
- C3 has one hydrogen atom.
According to Anti-Markovnikov addition:
- The H atom attaches to C2.
- The Br atom attaches to C3.
The product is: \( \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CH}(\text{Br}) - \text{CH}_3 \).
Naming from the right end to give the substituent (Bromo) the lowest number:
C1(\( \text{CH}_3 \)) - C2(\( \text{CH-Br} \)) - C3(\( \text{CH-CH}_3 \)) - C4(\( \text{CH}_3 \)).
The name is 2-Bromo-3-methylbutane.
Step 4: Final Answer:
The major product is 2-Bromo-3-methylbutane.